Question

One equivalent of calcium phosphate, $C{a}_3(P{O}_4{)}_2$ is dissolved in $HCl$.
$C{a}_3(P{O}_4{)}_2+6HCl\to 3CaC{l}_2+2H_{3}P{O}_4$
Moles of $CaC{l}_2$ formed are $\frac{1}{x}$. Then, the value of $x$ is:

Answer 1

$C{a}_3(P{O}_4{)}_2\rightleftharpoons 3C{a}^{2+}+2P{O}_4^{3-}$

Based on valency, equivalent weight of $C{a}_3(P{O}_4{)}_2=\frac{\text{Molecular weight}}{6}$

Hence, one equivalent $C{a}_3(P{O}_4{)}_2=\frac{\text{No. of moles}}{6}$

$1$ mole $C{a}_3(P{O}_4{)}_2=3$ moles $CaC{l}_2=2$ moles $H_{3}P{O}_4$

$\therefore \frac{1}{6}$ mol $C{a}_3(P{O}_4{)}_3=\frac{1}{2}$ mol $CaC{l}_2=\frac{1}{3}$ mol $H_{3}P{O}_4$

So, value of $x$ is $2$.